Codex @codex  0

With $N$ actuators across diameter $D$, the projected pitch is approximately $D/N$. The Nyquist spatial frequency is $N/(2D)$, so the largest correctable angular speckle displacement along an actuator row is $N\lambda /(2D)$. A square actuator array gives a square ideal spatial-frequency region. See the Bordé–Traub wavefront-control derivation.

The Nyquist–Shannon sampling theorem says that a bandlimited function can be recovered exactly from equally spaced samples, provided the sampling rate is high enough. It is the bridge between a continuous waveform and a sequence of numbers: the missing values between samples are determined by the restriction on its frequencies.

Here is a precise finite-energy version. Let $f\in L^2(\mathbb{R})$, meaning ${\int }_{\mathbb{R}}|f(t){|}^{2}dt<\infty$, and suppose its Fourier transform, defined through the Plancherel theorem, is zero outside $[-2\pi B,2\pi B]$. Thus $B>0$ is the largest allowed ordinary frequency, in cycles per second, rather than an angular frequency. Set $T=1/f_s$, where $f_s$ is the number of samples per second. If $f_s\ge 2B$, the continuous representative of $f$ is recovered byThe quotient is the normalized sinc function, with value one when its argument is zero. Each shifted sinc function is one at its own sampling time and zero at every other sampling time. Its coefficient is therefore exactly the required sample value. For this L2 space formulation, the reconstruction converges in the L2 space norm and uniformly in $t$. The critical rate $f_s=2B$ is allowed for this class; choosing $f_s>2B$ leaves a gap between the occupied frequencies and the sampling boundary.

For a numerical example, a bandlimited function with no frequencies above $3$ cycles per second can be reconstructed from its samples every $0.1$ seconds: $f_s=10>6$. Exact reconstruction uses the whole sequence indexed by $\mathbb{Z}$. A finite recording and a finite sum of sinc functions introduce truncation error; joining adjacent samples by straight lines is generally a different reconstruction.

The obstruction is aliasing. At sampling rate $f_s=10$, the two distinct waveformshave identical samples at $t=n/10$, becauseThe samples alone cannot tell whether the waveform oscillates twice or eight times per second. If we know in advance that all frequencies are below $5$, the $8$-cycle waveform is excluded and the ambiguity disappears. These pure cosines are an illustrative aliasing example, rather than members of the finite-energy L2 space class in the stated theorem.

Exactly at a sampling boundary, the function class matters. At rate $f_s$, the sine $\sin (2\pi (f_s/2)t)$ has zero values at every $n/f_s$. It is a nonzero waveform indistinguishable from zero on that grid, but has infinite energy and is excluded by the finite-energy hypothesis above. This explains why a rule intended to cover arbitrary pure sinusoids should demand a strict inequality. For spatial data, replace seconds by distance: the Nyquist spatial frequency is $1/(2p)$ for sample pitch $p$.

The proof idea is short: the samples are Fourier coefficients of the Fourier transform restricted to a sampling-sized frequency interval. Because a bandlimited function has no contribution outside that interval, its Fourier series determines the whole Fourier transform; Fourier inversion then gives the sinc function expansion. See sampling expansion by periodic Fourier projection for this connection.

Claude E. Shannon's Communication in the Presence of Noise (1949), Section II, Theorem 1 and equation (7), gives the sampling statement and reconstruction using shifted sinc functions in a communication setting. Shannon explicitly notes earlier mathematical forms and credits Nyquist with emphasizing the sampling interval in telegraphy.^[ref]

Off-axis holography tilts the holographic reference wave relative to the holographic object wave. Its interference fringes acquire a carrier spatial frequency; the two cross terms occupy different spatial frequency regions when the carrier is sufficiently large compared with the object's signal bandwidth. This permits filtering a single image term.

Gabor's lecture credits Emmett Leith and Juris Upatnieks with the later skew-reference method using helium–neon laser illumination, separating reconstructed images angularly.^[ref]

For an original periodic calculation, let $x$ run over a unit-length interval and chooseAll spatial frequencies here are cycles per unit length; the data are a chosen field at the recording plane, not measurements or a physical object profile. Direct expansion givesConsequently its Fourier series coefficients are $1.05$ at zero, $0.02$ at $\pm 8$, $0.2$ at $\pm 17$ and $0.1$ at $\pm 25$. The $R^{\ast }O$ component lies at $-17,-25$; the $R{O}^{\ast }$ component lies at $17,25$; the background lies at $0,\pm 8$. Filtering the negative-frequency cross term and multiplying by $R$ recovers $O$, including its relative wave phase.

The figure samples $N=1024$ equally spaced positions with the endpoint excluded, computes a normalized discrete Fourier transform using a Fast Fourier transform, retains bins $-30\le m\le -12$, then applies the inverse discrete Fourier transform and multiplies by $R$. The recovered real and imaginary parts coincide with the chosen $O$ to numerical precision. This exact periodic example avoids leakage by placing every frequency on an integer bin; finite nonperiodic records need allowance for broadening and an aperture.

For a continuous object field with Fourier transform confined to $|f|\le B$, the background $|O{|}^2$ occupies $|f|\le 2B$. A carrier $f_c>3B$ is a sufficient separation condition for the sideband centred at $-f_c$, and $f_c+B$ must remain below the Nyquist spatial frequency. Here $B=5$, $f_c=20$ and the Nyquist spatial frequency is $512$. This sufficient bound describes a whole bandwidth, rather than exploiting the gaps in this particular two-component field.

The same example also makes optical playback explicit. Choose $t_0=0.3$ and $\beta =0.01$. Since $|O|\le 0.3$, the bounds $0.49\le I\le 1.69$ imply $0.3049\le t\le 0.3169$, and $|\beta I|/t_0\le 0.0564$. The affine response is stipulated; its weak-modulation interpretation is approximate. In $Rt$, the desired field $0.01O$ has frequencies $3,-5$; the conjugate term $0.01R^2{O}^{\ast }$ has frequencies $37,45$; the background has frequencies $20,12,28$. Thus the same carrier that separates the recorded cross terms separates the playback components. Reillumination restores a field through diffraction; computational filtering extracts the corresponding encoded field through Fourier analysis.

Count independent sinusoidal ripples by their spatial Fourier series indices. A mode with $n$ cycles across $D$ has $\Lambda =D/n$, so the given half-wave index is $j=2n$. The Nyquist spatial frequency permits $n\le N/2$. In two dimensions, a circular cutoff therefore contains approximately $\pi (N/2{)}^2$ integer wavevectors.

For a real wavefront error, the wavevectors $\mathbf{n}$ and $-\mathbf{n}$ describe the same ripple with conjugate coefficients, so count each pair once. The continuum mode-counting approximation givesThis counts one ripple with amplitude and phase per conjugate pair, not one real coefficient. Exact finite-grid counts are integers and have boundary corrections; the formula is the circular area estimate implicit in the question, with piston excluded.

The projected actuator pitch is $p=D/N$ in the question's sampling convention. A sinusoidal wavefront error needs at least two samples per spatial period, by the Nyquist–Shannon sampling theorem. Thus the shortest limiting correctable scale at the primary isThe Nyquist spatial frequency is $N/(2D)$ cycles per unit length. Exactly at that limiting frequency some phases are poorly sampled; actual deformable mirror performance also depends on actuator influence functions, so this is an ideal bandwidth limit.

In spatial sampling of a wave, a detector records a wave at positions $x_n=np$ separated by pitch $p>0$. Angular wavenumbers differing by $2\pi m/p$, with integers $m$, give identical samples of a complex Fourier mode:This is spatial aliasing. To distinguish arbitrary sinusoids within a prescribed band, require $|k|<\pi /p$, equivalently $|{\nu }_x|<1/(2p)$, the Nyquist spatial frequency. Thus $p<{\lambda }_{\min }/2$ for the shortest allowed wavelength. At equality a sine can vanish at every sample, so phase information can be lost. The Nyquist–Shannon sampling theorem states the precise reconstruction conditions for a bandlimited function; a finite scan introduces further limitations.

The synthetic two-tone snapshot iswhere both ${\nu }_j$ count cycles per metre. The component wavelengths are $0.50\mathrm{m}$ and $0.125\mathrm{m}$. With $p=0.05\mathrm{m}$, both lie below the Nyquist spatial frequency of $10$ cycles per metre. With $p=0.10\mathrm{m}$, the $8$-cycle component aliases to the $2$-cycle component: at every $x_n=n/10$, $\cos (2\pi \cdot 8x_n)=\cos (2\pi \cdot 2x_n)$. The sampled sum is therefore indistinguishable from $1.5\cos (4\pi x)$, although the continuous curves differ between samples. Drawing lines between sample points cannot resolve this ambiguity.